View of Logic TK: Algebraic Notions from Tarski’s Consequence Operator

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São Paulo State University

Abstract. Tarski presented his definition of consequence operator to explain the most im- portant notions which any logical consequence concept must contemplate. A Tarski space is a pair constituted by a nonempty set and a consequence operator. This structure character- izes an almost topological space. This paper presents an algebraic view of the Tarski spaces and introduces a modal propositional logic which has as a model exactly the closed sets of a Tarski space.

Keywords: Tarski space, almost topological space, consequence operator, modal logic, alge- braic model.

Introduction

This work is inserted into the tradition of the algebraic logic, particularly, in the Lindenbaum-Tarski style.

The contribution of this paper to the algebraic logic field is to present the con- cept of Tarski’s consequence operator in an algebraic structure, the TK-algebra, and as well as to introduce a subnormal modal logic whose algebraic models are the counterpart of Tarski’s consequence operator.

Thus, a new propositional logic is generated and its adequacy in relation to TK- algebras is shown.

1. Tarski spaces

As follows, we adopt the concept of consequence operator in a slightly more general way than it was introduced by Tarski, in 1935.

Definition 1.1. A consequence operatoron E is a function⁻ :P(E) → P(E)such that, for everyA, B∈ P(E):

(i) A⊆A;

(ii) A⊆B⇒A⊆B;

Principia14(1): 47–70 (2010).

(iii) A⊆A.

From (i) and (iii), it follows thatA=Aholds, for everyA⊆E.

Definition 1.2. A consequence operator ⁻ : P(E) → P(E) is finitary when, for everyA⊆E:

A=∪{A₀:A₀ is a finite subset ofA}.

Definition 1.3. ATarski space (Tarski’s deductive system or Closure space) is a pair (E,⁻)such thatEis a nonempty set and⁻is a consequence operator on E.

Definition 1.4. Let(E,⁻)be a Tarski space. The setAisclosedin(E,⁻)whenA=A, andAisopenwhen its complement relative toE, denoted byA^C, is closed in(E,⁻). Proposition 1.5. In(E,⁻)any intersection of closed sets is also a closed set.

Proof. If{A_i}is a collection of closed sets, then∩iA_i ⊆ ∩iA_i ⊆ ∩iA_i =∩iA_i. Hence,

∩iA_i=∩iA_i.

Clearly,∅andEcorrespond to the least and the greatest closed sets, respectively, associated to the consequence operator⁻.

Proposition 1.6. In a structure of sets, the following conditions are equivalent:

(i) A⊆B⇒A⊆B;

(ii) A⊆A∪B.

Proof. (i)⇒(ii): AsA⊆A∪B, by (i), it follows thatA⊆A∪B. (ii)⇒(i): IfA⊆B, thenA∪B=B. So, by (ii),A⊆B.

Definition 1.7. A Tarski space(E,⁻)isvacuouswhen∅=∅.

Definition 1.8. Analmost topological spaceis a pair(S,Ω)such thatSis a nonempty set andΩ⊆ P(S)satisfies the following condition:

B⊆Ω⇒ ∪B∈Ω.

The collectionΩis calledalmost topologyand each member ofΩis anopenof(S,Ω). A setA∈ P(S)isclosedwhen its complement relative toS is an open of(S,Ω). Proposition 1.9. In an almost topological space(S,Ω)the set∅is open and S is closed.

Proposition 1.10. In an almost topological space(S,Ω), any intersection of closed sets is still a closed set.

Definition 1.11. Let (S,Ω)be an almost topological space. Theclosure ofAis the set:

A=df∩{X ⊆S:X is closed andA⊆X}.

Proposition 1.12. Let(S,Ω) be an almost topological space. For every A⊆ S, A is closed.

Proposition 1.13. Let (S,Ω)be an almost topological space and let A be defined as above, for every A⊆S. Then(S,⁻)is a Tarski space.

On the other hand, if (E,⁻) is a Tarski space, let us consider Ω = {X ⊆ E : X is open}.

Proposition 1.14. If(E,⁻)is a Tarski space, then(E,Ω)is an almost topological space.

It follows from Propositions 1.13 and 1.14 that that given a Tarski space an almost topological space is obtained and in another direction given an almost topo- logical space we can define a Tarski space. So naturally we can make an interrelation between the two concepts.

Definition 1.15. An almost topological space(S,Ω)is0-closedwhen it holds:

(iv) ∅=∅.

Definition 1.16. A topological space (S,Ω)is an almost topological space 0-closed such that it holds:

(v) A∪B=A∪B.

The previous definition of topological space was given by the Kuratowski’s clo- sure. Naturally, every topological space is a Tarski space, but there are several Tarski spaces which are not topological spaces. Each topological space is an instance of a vacuous Tarski space.

2. TK-algebras

The definition of a TK-algebra introduces the notions of consequence operator in the context of the algebraic structures.

Definition 2.1. ATK-algebra is a sextupleA = (A, 0, 1,∨,∼,•)such that(A, 0, 1,∨,

∼)is a Boolean algebra and•is a new operator, calledoperator of Tarski, such that:

(i) a∨ •a=•a;

(ii) •a∨ •(a∨b) =•(a∨b);

(iii) •(•a) =•a.

Examples:

(a) The space of setsP(A)withA6=∅and•a=a, for alla∈A, is a TK-algebra.

(b) The space of setsP(R)with•X =X∪ {0}is a TK-algebra.

(c) The space of sets P(R) with •X = ∩{I | I is an interval andX ⊆ I} is a TK- algebra.

Since we are working with Boolean algebras, the item (i) of the Definition 2.1 asserts that, for everya∈A,a≤ •a.

We define in a TK-algebra:

ašb =df ∼a∨b;

a−b =df a∧ ∼b.

Proposition 2.2. In any TK-algebra the following conditions are valid:

(i) ∼ •a≤ ∼a≤ • ∼a;

(ii) a≤ b⇒ •a≤ •b.

Proof. (ii)a≤b⇒a∨b=b⇒ •(a∨b) =•b⇒ •a∨•b=•a∨•(a∨b) =•(a∨b) =

•b⇒ •a≤ •b.

Proposition 2.3. In any TK-algebra the following assertions are valid:

(i) •(a∧b)≤ •a∧ •b;

(ii) •a∨ •b≤ •(a∨b).

Proof. (i)a∧b≤aanda∧b≤b⇒ •(a∧b)≤ •aand•(a∧b)≤ •b⇒ •(a∧b)≤

•a∧ •b.

(ii) is similar to (i).

Example:

(a) Let E={a, b, c}. The space of sets(P(E),∅,E,∩,∪,^C) can be extended to a TK-algebra in the following way: •X = X, for allX ⊆ E such thatX 6={a,b}, and

•{a,b}=E.

IfX ={a}andY ={b}, then•X=•{a}={a}and•Y =•{b}={b};•X∪ •Y = {a,b};•(X∪Y) =•{a,b}=E. Hence,•X∪ •Y ⊂ •(X∪Y).

IfX ={a, b}andY ={c}, then•X =•{a, b}=E,•Y =•{c}={c},•X∩ •Y = {c},•(X∩Y) =•∅=∅. Hence,•(X∩Y)⊂ •X∩ •Y.

Proposition 2.4. In any TK-algebra, it holds:

(i) •(•a∧ •b) =•a∧ •b;

(ii) •(•a∨ •b) =•(a∨b); (iii) •aš•b≤ •(ašb).

Proof. (i) It is enough to verify that •(•a∧ •b) ≤ •a∧ •b. But, •(•a∧ •b) ≤

• •a∧ • •b=•a∧ •b.

(ii)•(a∨b)≤ •(•a∨ •b)≤ • •(a∨b) =•(a∨b). Hence,•(a∨b) =•(•a∨ •b). (iii)•aš•b=∼ •a∨ •b≤ • ∼a∨ •b≤ •(∼a∨b)=•(ašb).

We have a new operation in a TK-algebra, dual of•:

◦a=df∼ • ∼a.

Proposition 2.5. In a TK-algebra, the following conditions are valid:

(i) ◦a¶a;

(ii) a≤ b⇒ ◦a≤ ◦b;

(iii) ◦(a∧b)≤ ◦a;

(iv) ◦a≤ ◦ ◦a.

Proof. (i)◦a=∼ • ∼a⇒ ∼a≤ • ∼a=∼ ◦a⇒ ◦a≤ ∼ ∼a=a.

(ii)a≤ b⇒ ∼b≤ ∼a⇒ • ∼b≤ • ∼a⇒ ∼ • ∼a≤ ∼ • ∼b⇒ ◦a≤ ◦b.

(iii) It follows from (ii).

(iv)◦a≤a⇒ ∼a≤ ∼ ◦a⇒ • ∼a≤ • ∼ ◦a⇒ ∼ • ∼ ◦a≤ ∼ • ∼a⇒ ◦ ◦a≤ ◦a.

Definition 2.6. An element a ∈Aisclosed when•a= a, and a ∈Aisopen when

◦a=a.

Proposition 2.7. In any TK-algebra:

(i) If a is open, then: a≤b⇔a≤ ◦b;

(ii) If b is closed, then: a≤ b⇔ •a≤b.

Definition 2.8. An algebraA isnon-degeneratewhen its universeAhas at least two elements.

Definition 2.9. LetA = (A, 0, 1,∨,∼,•) andB = (B, 0, 1,∨,∼,•) be TK-algebras.

A homomorphism between A and B is a function h: A−→ B that preserves the TK-operations. Thekernelofhis the setKer(h) =df{x ∈A:h(x) =0}=h⁻¹(0). Theorem 2.10. For each TK-algebraA = (A, 0, 1,∨,∼,•)there is a monomorphism h from A into a Tarski space of sets defined inP(P(A)).

Proof. Through Stone’s isomorphism, we know that for each Boolean algebraA = (A, 0, 1,∨,∼)there is a monomorphismhfromAinto a field of subsets ofP(A).

Next, we introduce a Tarski space inP(A)in the following way:

For each set X⊆ P(A), we define:

X =∩a∈A{h(a):X⊆h(a)anda=•a}. Then, we must show that:

(i) X ⊆X

(ii) X ⊆Y ⇒X ⊆Y (iii) X ⊆X

(iv) h(•a) =h(a).

We can see that (i) - (iv) are valid by:

(i) By definition ofX.

(ii) Suppose X * Y. Then there is z such that z ∈X and z ∈/ Y. So, for some a∈A, z ∈/ h(a)withY ⊆h(a)and•a= a. Since X ⊆ Y ⊆h(a) and•a =a, then z∈/X, and it contradictsz∈X.

(iii) x ∈X ⇒ x ∈ ∩_a∈A{h(a):X ⊆h(a)anda =•a}. AsX ⊆X, it follows that x∈ ∩a∈A{h(a):X ⊆h(a)anda=•a}=X and, therefore,X ⊆X .

(iv) On one hand,h(•a)⊆h(a)is valid: Consider thath(a)⊆h(b)and b=•b.

Since h is a Boolean monomorphism, a ≤ b and •a ≤ •b. But, since •b = b, then•a ≤ b andh(•a)⊆h(b). Concluding, for each b∈Asuch that•b= b and h(a)⊆h(b), it results thath(•a)⊆h(b), that is,h(•a)⊆h(a).

On the other hand, h(a) ⊆ h(•a) is also valid: a ≤ •a ⇒ h(a) ⊆ h(•a) ⇒ h(a)⊆h(•a). Sinceh(•a)⊆h(•a)and•a=• •a, thenh(•a) =h(•a)andh(a)⊆ h(•a).

3. Ideals in TK-algebras

As in Boolean algebras, we can define an ideal into TK-algebras and use it to analyze aspects of the consequence in the next sections.

Definition 3.1. LetA = (A, 0, 1,∨,∼,•)be a TK-algebra. AnidealinAis a nonempty set I⊆Asuch that, for all x,y ∈A:

(i) x,y∈I⇒x∨y ∈I;

(ii) x ∈I and y≤x ⇒y ∈I.

If I is an ideal anda₁,a₂,· · ·,a_n ∈I, by induction on n, we have thata₁∨a₂∨

· · · ∨a_n∈I.

Definition 3.2. The idealI is aTK-idealwhenx ∈I⇒ •x∈I. Examples:

(a) The setAis a TK-Ideal inA.

(b) The single{0}is a TK-ideal if, and only if,•0=0.

(c) Given a ∈A, the set [a] ={x ∈A: x ≤ •a}is a TK-Ideal. The set [a]is the TK-Idealgeneratedbya.

Proposition 3.3. Let A be a TK-algebra and I be an ideal in A. The following conditions are equivalent:

(i) a∈I⇒ •a∈I ;

(ii) ašb∈I⇒ •aš•b∈I .

Proof. (⇒)ašb∈I⇒ •(ašb)∈I. Since•aš•b≤ •(ašb), then•aš•b∈ I.

(⇐) Ifa∈I, asa=1ša, then 1ša∈I. By (ii),•1š•a∈I ⇒ 1š•a∈I⇒

•a∈I.

Proposition 3.4. Let A be a TK-algebra and ∅ 6= B ⊆ A. The set [B] ={x ∈ A: (∃a₁, . . . ,a_n∈B)(x ≤ •(a₁∨. . . ∨a_n)}is a TK-ideal.

Proof. (i) If x,y ∈ [B], then there are a₁, . . . ,a_n,b₁, . . . ,b_n ∈ B such that x ≤

•(a₁∨. . .∨a_n)andy ≤ •(b₁∨. . .∨b_n). Sox∨y ≤ •(a₁∨. . . ∨a_n)∨•(b₁∨. . .∨b_n)≤

•(a1∨. . . ∨a_n∨b₁∨. . . ∨b_n) and x ∨y ∈[B]. (ii) If x ∈[B]and y ≤ x, then y ∈[B]. (iii) Ifx∈[B], then there existsa₁, . . . ,a_n∈Bsuch thatx ≤ •(a₁∨. . .∨a_n) and since •(a1∨. . . ∨a_n)is closed, by Proposition 2.7,•x ≤ •(a1∨. . . ∨a_n) and

•x∈[B].

Definition 3.5. TK-ideal[B]defined in the Proposition 3.4 is the TK-idealgenerated by B.

Proposition 3.6. LetA be a TK-algebra. If I is a TK-Ideal inA and b∈A, then [I,b] ={x ∈A:(∃c∈I)(•x≤ •(b∨c))}

is a TK-Ideal.

Proof. Of course, if x ∈[I,b]and y ≤ x, then y ∈[I,b]; and x ∈[I,b]⇒ •x ∈ [I,b].

Now, ifx,y∈[I,b], there arec,d ∈Isuch that•x ≤ •(b∨c)and•y ≤ •(b∨d). Hence, •(x∨ y) =•(•x∨ •y) ≤ •(•(b∨c)∨ •(b∨d)) =•((b∨c)∨(b∨d)) =

•(b∨(c∨d)). Therefore, x∨y∈[I,b].

Definition 3.7. The ideal[I,b]is the TK-idealgeneratedbyI and b.

Definition 3.8. Let I be an ideal in a TK-algebra A. The ideal I isproper when I 6=A. The ideal I ismaximalwhen it is proper and it is not included in any proper ideal distinct of I. The idealI isprimewhen it is proper and for alla,b∈Ait holds:

a∧b∈I⇒a∈I or b∈I.

Of course, if I is proper, then there is an a ∈Asuch that a ∈/ I and therefore 1∈/ I.

Proposition 3.9. Let I be an ideal in a TK-algebra A. The following statements are equivalent:

(i) I is maximal;

(ii) for every a∈A: a∈I Ù∼a∈I ; (iii) I is prime;

(iv) for every a,b∈A, a−b∈I or b−a∈I .

Proof. As A is a Boolean algebra (i), (ii) and (iii) are equivalent. Let’s show the equivalence between (iii) and (iv). (iii)⇒(iv) Let a,b∈A. As(a−b)∧(b−a) = 0 ∈ I and I is prime, then a−b ∈ I or b−a ∈ I. (iv) ⇒ (iii) Let a∧b ∈ I. If a∧ ∼b= a−b∈I, then a = (a∧b)∨(a∧ ∼b)∈I. If b∧ ∼a= b−a∈ I, then b∈I.

The next definitions are specific for TK-ideals.

Definition 3.10. A TK-Ideal I is TK-irreducible when it is proper and for any two TK-IdealsI₁andI₂:

I=I₁∩I₂⇒I=I₁ or I=I₂.

Definition 3.11. A TK-IdealIisTK-maximalwhen it is proper and it is not included in any proper TK-Ideal distinct ofI.

Definition 3.12. Let I be a TK-ideal in a TK-algebra A. The ideal I is TK-prime when it is a proper ideal and for alla,b∈Ait holds:

•a∧ •b∈I⇒ •a∈I or•b∈I.

Let I be a TK-ideal. If I is a prime ideal then I is a TK-prime ideal. However, it is possible that I is a TK-prime ideal that is not prime. The same holds to TK- maximal ideals. Besides, being maximal does not imply to be TK-ideal and hence to be TK-maximal ideal.

Example:

(a) Let E = {a,b,c}. Consider the TK-algebraA = (P(E),∅,E,∩,∪, ^C,•), such that •∅ = ∅,•{a}= {a},•{b} =•{c}= •{a,b}= •{a,c}= •{b,c} = •{a,b,c}= {a,b,c}. The TK-ideal{∅,{a}}is TK-prime but it is not prime.

Proposition 3.13. If a TK-ideal is TK-maximal, then it is TK-irreducible.

Proof. LetI be a TK-maximal ideal. So I is proper. Now, if I₁ andI₂ are two proper TK-ideals such that I = I₁∩I₂, then I ⊆ I₁ and I ⊆ I₂. As I is TK-maximal, so I=I₁=I₂.

Example:

(a) Let E={a,b}. Consider the TK-algebraA =(P(E),∅,E,∩,∪, ^C,•), such that

•∅=∅,•{a}={a},•{b}=•{a,b}={a,b}. The TK-ideal{∅}is TK-irreducible and it is contained in the TK-ideal maximal{∅,{a}}.

It follows from the previous example that we have a TK-irreducible ideal which is not TK-maximal.

Proposition 3.14. Let A be a TK-algebra and I a TK-ideal in A. The following statements are equivalent:

(i) I is a TK-prime ideal;

(ii) for every a∈A: either•a∈I or∼ •a∈I .

Proof. (i)⇒(ii) If Iis TK-prime, then I is proper. Now, if•a∈I and∼ •a∈I, then

•a∨ ∼ •a=1∈I, which is a contradiction. Besides, asI is prime and•a∧ ∼ •a= 0∈I, then•a∈I or∼ •a∈I.

(ii)⇒(i) By hypothesis,I is proper. If•a∧ •b∈I and•a∈/ I, then, by hypothesis,

∼ •a∈I. Hence,•b≤ •b ∨∼ •a=1∧(•b ∨∼ •a) = (•a ∨∼ •a)∧(•b ∨∼ •a)= (•a∧ •b) ∨ ∼ •a∈I. SinceI is a TK-ideal, then•b∈I. Therefore,I is a TK-prime ideal.

Proposition 3.15. If a TK-ideal is TK-prime, then it is TK-maximal.

Proof. ConsiderIas a TK-prime ideal which is properly included in a TK idealM and take x such thatx ∈M, butx ∈/ I. SinceI is a TK-ideal, then•x ∈/I and sinceM is a TK-ideal,•x ∈M. AsI is a TK-prime ideal, by Proposition 3.14,∼ •x ∈I. Hence,

•x∈M,∼ •x∈I⊆M, therefore 1=•x∨ ∼ •x∈M, that is,M=A.

Corollary 3.16. If an TK-ideal is TK-prime, then it is TK-irreducible.

Proof. It follows from Propositions 3.15 and 3.13.

Example:

(a) LetE={a,b,c}. Consider the TK-algebraA =(P(E),∅,E,∩,∪, ^C,•), such that

•∅=∅,•{a}={a},•{b}={b},•{c}={c}, •{a,b}={a, b},•{a, c}=•{b, c}=

•{a, b, c}= {a, b, c}. The TK-ideal I ={∅,{c}} is TK-maximal, but it is not TK- prime:

(i) I is TK-maximal: if I ⊂J with J a TK-ideal, thenJ has an element x such that x ∈/ I. If x = {a}, then {a,c} = {a} ∪ {c} ∈ J, and like J is a TK-ideal, then {a,b,c}=•{a,c} ∈J and therefore J=A. If x ={b}or x ={a,b}or x ={a,c}or

x={b,c}or even x={a,b,c}, we show that J=A.

(ii) I is not TK-prime: •{a} ∧ •{b}=∅∈I, but•{a}∈/I and•{b}∈/I.

Definition 3.17. Achain of ideals is a sequence (I₁,I₂,I₃, . . . ) of ideals such that I₁⊆I₂⊆I₃⊆. . . .

Lemma 3.18. LetA be a TK-algebra and(I₁,I₂,I₃, . . .)be a chain of proper TK-ideals ofA. The union∪I_n is a proper TK-ideal.

Proof. Let x ∈ I and y ∈ A, with y ≤ x. Since I = ∪I_n, there is n ∈N such that x ∈ I_n, and since I_n is a TK-ideal, then y ∈ I_n ⊆ I. Let x,y ∈ I = ∪I_n. Since I₁ ⊆ I₂ ⊆ I₃ ⊆ . . . , there is n∈N such that x,y ∈ I_n, and as I_n is a TK-ideal, so x∨y∈I_n⊆I. Hence,I is an ideal. Since 1∈/I_n, for everyn, hence 1∈/I. Therefore, I is a proper ideal. It is immediate thatI is a TK-ideal.

Theorem 3.19. Each proper TK-ideal in a TK-algebra is contained in a TK-maximal TK-ideal.

Proof. The result follows from the previous lemma and by Zorn’s Lemma.

Corollary 3.20. LetA be a TK-algebra and a∈A such that•a6=1. Then, there is a maximal TK-ideal I for which a∈I .

Proof. Since•a6=1, the TK-ideal generated byais proper, then by previous theorem, [a]is included in a TK-maximal TK-idealI anda∈I.

Proposition 3.21. LetA be a TK-algebra and I be a TK-ideal in A. If a∈/ I , then there is a TK-irreducible TK-ideal I^∗such that I⊆I^∗and a∈/ I^∗.

Proof. LetS be the ordered set of all TK-ideals J ofA such thatI⊆J anda∈/J. By Zorn’s Lemma and Lemma 3.18, there is a TK-maximal elementM inS. IfM =I₁∩I₂, considering thata∈/M, thena∈/ I₁ora∈/ I₂. Ifa∈/I₁, sinceI⊆M⊆I₁, thenI₁∈S and as M is TK-maximal in S, M = I₁. Besides, if a ∈/ I₂, M = I₂. Hence, M is TK-irreducible.

Theorem 3.22. Let h:A−→B be a surjective homomorphism betweenA = (A, 0, 1,∨,

∼,•)andB = (B, 0, 1,∨,∼,•). Then:

(i) h(a) =h(b)⇔a≡_Ker(h)b⇔d f a−b∈Ker(h)and b−a∈Ker(h); (ii) the relation≡Ker(h)is a congruence;

(iii) A |_Ker(h)is a TK-algebra;

(iv) A |Ker(h)tB.

Proof. (i) (⇒)h(a) = h(b) ⇒ h(a)∧ ∼h(b) = 0 ⇒ h(a∧ ∼b) =0 ⇒ a∧ ∼b ∈ ker(h)⇒a−b∈ker(h). In the same way, b−a∈ker(h). (⇐)a−b∈ker(h)⇒ a∧ ∼b∈ker(h) ⇒h(a∧ ∼b) =0⇒h(a)∧ ∼h(b) =0⇒h(a)∨h(b) =h(b)⇒ h(a)≤h(b). Also, b−a∈ker(h)⇒h(b)≤h(a). So,h(a) =h(b).

(ii) Clearly≡Ker(h)is an equivalence relation. We only need to show the details about how to extend to a TK-algebra, that is, a≡_Ker(h) b⇒ •a≡_Ker(h)•b. Soa≡_Ker(h) b

⇔h(a) =h(b)⇒ •h(a) =•h(b)⇒h(•a) =h(•b)⇔ •a≡Ker(h)•b.

(iii) Since ≡_Ker(h) is an equivalence we have a partition on A |_Ker(h). Now, let⁻ : A −→ A |Ker(h) such that a = {b ∈ A: a ≡Ker(h) b}. We must produce a Tarski operator inA |_Ker(h). Let•a=•a. This is the looked operator: (1)a∨•a=a∨ •a=

•a = •a; (2) •a∨ •(a∨b) = •a∨ •(a∨b) = •a∨ •(a∨b) = •a∨ •(a∨b) =

•(a∨b) =•a∨b=•(a∨b); (3)• •a=• •a=• •a=•a=•a.

(iv) As usual, we define ˜h: A |Ker(h) −→ B by ˜h(a) = h(a). Naturally ˜h is well defined and bijective. We will show that ˜h preserves the operation •: ˜h(•a) =

˜h(•a) =h(•a) =•h(a) =•˜h(a).

Definition 3.23. LetA be a TK-algebra andI a maximal (or prime) TK-ideal inA. Consider the following (equivalence) relation≡I inA:

a≡I b⇔=dfa−b∈I andb−a∈I.

In the above definition,I more than TK-maximal is maximal.

The next theorem will show that the relation ≡I is a congruence in A with respect to∧,∨,∼and•, that is,[a] = [b]implies[∼a] = [∼b]and[•a] = [•b]; and[a] = [b],[c] = [d]implies[a∧c] = [b∧d]and[a∨c] = [b∨d]. Also,[1]is the unit,[0]is the zero element ofA |I.

Proposition 3.24. LetA be a TK-algebra and I a maximal TK-ideal inA. (i) the relation≡I determined by I is a congruence relation;

(ii) considering, for every x ∈A, •[x] = [•x], the quotient algebra A |I is a TK- algebra;

(iii) the function h:A −→ A |I defined by h(a) =a is a surjective homomorphism;

(iv) I is the kernel of h, and for every a∈A, a∈I⇔a≡I0;

(v) for every a∈A, a∈/I ⇔a≡I1.

Proof. (i) Clearly≡I is an equivalence relation. Now:

(1) [a] = [b]⇔ a−b ∈ I and b−a ∈ I ⇔ a ∧ ∼b ∈ I and b ∧ ∼a ∈ I ⇔

∼a ∧b∈I and∼b∧a∈I ⇔ ∼a− ∼b∈Iand∼b− ∼a∈I ⇔[∼a] = [∼b]. (2)[a] = [b]⇔a−b∈I and b−a∈I ⇔a ∧ ∼b∈I and b ∧ ∼a∈I. Ifa∈I, then •a ∈ I and∼a ∈/ I ⇒ ∼ •a ∈/ I and b ∈ I ⇒ •b ∈ I ⇒ •a − •b ∈ I and

•b − •a ∈I ⇔[•a] = [•b]. Ifa ∈/ I, then•a∈/ I and∼b∈I ⇒ ∼ •a ∈I and b∈/ I ⇒ •b∈/ I ⇒ ∼ •b∈I ⇒ •a − •b∈I and•b − •a∈I ⇔[•a] = [•b]. In any case, we have[•a] = [•b].

(3) If[a] = [b]and[c] = [d], thena∧∼b∈I,b∧∼a∈I,c∧∼d∈Iandd∧∼c∈I.

SinceIis an ideal, anda∧ ∼b,c∧ ∼d∈I, thena∧c∧ ∼b,a∧c∧ ∼d∈I, therefore (a∧c)∧ ∼(b∧d) = (a∧c)∧(∼b∨ ∼d) = (a∧c∧ ∼b)∨(a∧c∧ ∼d)∈I. That is, (a∧c)−(b∧d)∈I. Analogously,(b∧d)−(a∧c)∈I. Hence,[a∧c] = [b∧d]. (4) If[a] = [b]and[c] = [d]thena∧∼b∈I,b∧∼a∈I,c∧∼d∈Iandd∧∼c∈I.

Since I is an ideal, anda∧ ∼b, c∧ ∼d ∈ I, thena∧ ∼b∧ ∼d, c∧ ∼b∧ ∼d ∈ I, therefore(a∨c)∧∼(b∨d) = (a∨c)∧(∼b∧∼d) = (a∧∼b∧∼d)∨(c∧∼b∧∼d)∈I.

That is,(a∨c)−(b∨d)∈I. Analogously,(b∨d)−(a∨c)∈I. Hence,[a∨c] = [b∨d]. (ii) We need to verify that• |_≡I preserves the properties of operator•.

(1)[x]∨ •[x]=[x]∨[•x]=[x∨ •x]=[•x] =•[x].

(2)•[x]∨ •[x∨y]=[•x]∨[•(x∨y)]=[•x∨ •(x∨y)]=[•(x∨y)] =•[(x∨y)]. (3)• •[x] = [• •x] = [•x] =•[x].

(iii) It is immediate.

(iv) (⇒) Ifa∈I, asa=a∧1=a∧ ∼0=a−0∈I and 0=0∧ ∼a=0−a∈I.

Soa≡I0. (⇐) Ifa≡I0 thena=a∧1=a∧ ∼0=a−0∈I.

(v)a∈/I ⇔ ∼a∈I ⇔ ∼a≡I0⇔a≡I1.

We observe that ifI is an ideal thenA |I is degenerate⇔I=A.

Proposition 3.25. If I is a prime TK-ideal, then the quotient algebraA |_≡I is linearly ordered.

Proof. LetIbe a prime TK-ideal and takea,b∈A. From this we have thata−b∈Ior b−a∈I, that is,[a]≤[b]or[b]≤[a]and, thereforeA |I is linearly ordered.

In the next section, a new logic associated with Tarski’s consequence operator is introduced.

4. Logic TK

The propositional logic TK is the logical system associated to the TK-algebras. The propositional language of TK is L = (¬,∨,→,‡,p₁, p₂,p₃, . . .) and TK is presented as follows:

Axioms:

(CPC) ϕ, ifϕis a tautology;

(TK₁) ϕ→‡ϕ; (TK₂) ‡‡ϕ→‡ϕ. Deduction Rules:

(MP) ϕ→ψ, ϕ ψ ; (RM^‡) `ϕ→ψ

`‡ϕ→‡ψ.

As usual, we write`Sϕto indicate thatϕis a theorem of some axiomatic system S, and we drop the subscript when there is no possibility of misunderstanding.

Definition 4.1. LetΓbe a set of formulas, andϕ a formula of some systemS. We say that Γ deduces ϕ (notation: Γ `S ϕ) if there is a finite sequence of formulas ϕ1, . . . ,ϕn such thatϕn=ϕ and, for everyϕi, 1≤i≤n,

(i) ϕi is an axiom; or (ii) ϕi ∈Γ; or

(iii) ϕi is obtained from previous formulas of the sequence by some of the deduc- tion rules.

Notice that the notion of syntactic consequence (deduction) presented here is global. Accordingly, we will have, for instance, p→q`‡p→‡q. However,(p→ q) →(‡p → ‡q) is not a theorem (what we can show semantically after proving completeness), and so the Deduction Theorem does not hold¹.

Proposition 4.2. `‡ϕ→‡(ϕ∨ψ).

Proof.

1.ϕ→(ϕ∨ψ) Tautology

2.‡ϕ→‡(ϕ∨ψ) R^‡in 1

Proposition 4.3. `ϕ⇒ `‡ϕ.

Proof.

1.ϕ Premise

2.ϕ→‡ϕ Ax_TK1

3.‡ϕ MP in 1 and 2.

Proposition 4.4. Γ`‡ϕ∨‡ψ→‡(ϕ∨ψ) Proof.

1.‡ϕ→‡(ϕ∨ψ) Proposition 4.2

2.‡ψ→‡(ϕ∨ψ) Proposition 4.2

3.‡ϕ∨‡ψ→‡(ϕ∨ψ) CPC.

As in the case of a TK-algebra, we can define the dual operator of ‡ in the following way:

◊ϕ=df¬‡¬ϕ. Proposition 4.5. `ϕ→ψ⇒ `◊ϕ→◊ψ.

Corollary 4.6. `ϕ↔ψ⇒ `◊ϕ↔◊ψ.

Proposition 4.7. `◊ϕ→ϕ.

Proposition 4.8. `◊ϕ→◊◊ϕ.

Proposition 4.9. `◊(ϕ∧ψ)→◊ϕ.

Corollary 4.10. `◊(ϕ∧ψ)→(◊ϕ∧◊ψ).

We could, alternatively, consider the operator◊as primitive and substitute the axioms TK₁and TK₂for the following ones:

(TK^∗₁) ◊ϕ→ϕ, (TK^∗₂) ◊ϕ→◊◊ϕ,

and the rule RM^‡by the rule RM^◊: (RM^◊) ϕ→ψ

◊ϕ→◊ψ.

In the following section, the algebraic adequacy of the Logic TK relative to TK- algebras is shown.

5. The algebraic adequacy

Below, we indicate the set of propositional variables of TK by Var(TK), the set of its formulas by For(TK) and a generic TK-algebra by A. The propositional logical system TK is determined by a pair (L,⁻), where L is the propositional language of TK and⁻is a consequence operator on For(TK) given by axioms and deduction rules of TK.

Thus, forΓ ⊆For(TK), denoting the set of axioms of TK by Ax, thenΓ = {ψ: Γ∪Ax`ψ}. We say thatψis derivable in TK or is a theorem of TK whenψ∈∅, or, Γ =∅.

Definition 5.1. ATK-theoryis a set∆⊆For(TK), such that∆ = ∆. When∆ =∅, we have the theorems of TK, that is,ψ∈∅⇔ `ψ.

Definition 5.2. A formula ψ∈For(TK) isrefutableinΓwhenΓ` ¬ψ. Otherwise, ψisirrefutable.

Definition 5.3. A set Γ⊆ For(TK) is irreducibleif it is consistent and for any two sets∆1,∆2⊆For(TK):

Γ = ∆1∩∆2⇒Γ = ∆1 ∨ Γ = ∆2.

Definition 5.4. A setΓ⊆For(TK) ismaximalif it is consistent and for any consistent set∆⊆For(TK):

Γ⊆∆⇒Γ = ∆.

Definition 5.5. Arestrict valuation is a function ˘v: Var(TK) −→ A that interprets each variable of TK in an element ofA.

Definition 5.6. Avaluationis a functionv: For(TK)−→ A that extends natural and uniquely vas follows:

(i) v(p) =˘v(p); (ii) v(¬ϕ) =∼v(ϕ); (iii) v(‡ϕ) =•v(ϕ);

(iv) v(ϕ∧ψ) =v(ϕ)∧v(ψ); (v) v(ϕ∨ψ) =v(ϕ)∨v(ψ).

As usual, operator symbols of left members represent logical operators and the right ones represent algebraic operators.

Definition 5.7. LetA be a TK-algebra. A valuation v: For(TK) −→ A is a model for a setΓ⊆For(TK) whenv(γ) =1, for each formulaγ∈Γ.

In particular, a valuation v: For(TK)−→ A is a model for ϕ ∈For(TK) when v(ϕ) =1.

Definition 5.8. A formula ϕ is valid in a TK-algebra A when every valuation v:

For(TK)−→ A is a model forϕ.

Definition 5.9. A formulaϕ is TK-valid, what is denoted byϕ, when it is valid in every TK-algebra.

If we consider the set of formulas For(TK), naturally we have an algebra on For(TK), B = (For(T K),∧,∨,∼,‡)such that∧and∨are binary operators, ¬and

‡are unary operators.

Now, we define the Lindembaum algebra of TK.

Definition 5.10. LetΓ∪ {ϕ,ψ} ⊆For(TK) and' the equivalence relation defined by:

ϕ'ψ⇔d f Γ`ϕ→ψand Γ`ψ→ϕ.

The relation', more than an equivalence, is a congruence, since by the ruleR^‡: ϕ ' ψ⇒ Γ`ϕ →ψ and Γ`ψ →ϕ⇒ Γ`‡ϕ →‡ψ and Γ`‡ψ→‡ϕ⇒

‡ϕ'‡ψ.

For eachψ∈For(TK), we denote the class of equivalence ofψmodulo'andΓ by[ψ]_Γ={σ∈For(T K):σ'ψ}.

The Lindembaum algebra of TK, denoted by A_Γ(T K), is the quotient algebra B|_', defined by:

A_Γ(T K)=(For(T K)|_', 0, 1,¬_',‡_',∧_',∨_'), such that:

0= [ϕ∧ ¬ϕ], 1= [ϕ∨ ¬ϕ],

¬_'[ϕ] = [¬ϕ],

‡_'[ϕ] = [‡ϕ],

[ϕ]∧_'[ψ] = [ϕ∧ψ], [ϕ]∨_'[ψ] = [ϕ∨ψ].

In general, we do not indicate the index_' of operations. When Γ =∅ we just writeA(T K).

Proposition 5.11. InA_Γ(T K)it is valid: [ϕ]≤[ψ]⇔Γ`ϕ→ψ.

Proof. [ϕ]≤[ψ]⇔[ϕ]∨[ψ] = [ψ]⇔[ϕ∨ψ] = [ψ]⇔Γ`ϕ∨ψ↔ψ⇔ Γ`ϕ→ψ.

Proposition 5.12. The algebraA_Γ(T K)is a TK-algebra.

Proof. Ax_{T K1}:Γ`ϕ→‡ϕ⇒[ϕ]≤[‡ϕ]⇒[ϕ]≤‡[ϕ];

Proposition 4.2: Γ`‡ϕ→‡(ϕ∨ψ)⇒[‡ϕ]≤[‡(ϕ∨ψ)]⇒‡[ϕ]≤‡[ϕ∨ψ]; Ax<sub>T K2</sub>: Γ`‡‡ϕ→‡ϕ⇒[‡‡ϕ]≤[‡ϕ]⇒‡‡[ϕ]≤‡[ϕ].

Definition 5.13. The algebraAΓ(T K)is thecanonical modelofΓ⊆For(TK).

As an immediate consequence,A(T K)is the canonical model for the theorems of TK.

Corollary 5.14. LetΓ∪ {ϕ} ⊆For(TK):

(i) If Γ`ϕ, then[ϕ] =1inA_Γ(T K);

(ii) If Γ` ¬ϕ (ϕ is refutable inΓ), then[ϕ] =0inA_Γ(T K);

(iii) IfA_Γ(T K)is non degenerate, then there is a formula that is not a theorem of Γ.

Proof. SinceA_Γ(T K)always has an identity element 1, then for everyϕ∈For(TK), [ϕ]≤1.

(i) The formula ϕ →(ψ→ϕ)is a tautology and, hence, a theorem of TK. With a substitution we have`ϕ →((ϕ→ϕ)→ϕ). Now, if Γ`ϕ, by MP, it follows that Γ`(ϕ→ϕ)→ϕ, that is, 1= [ϕ→ϕ]≤[ϕ]and[ϕ] =1.

(ii) Letϕbe refutable inΓ, that is,Γ` ¬ϕ. ButΓ` ¬ϕiff[¬ϕ] =1 iff∼[ϕ] =1 iff [ϕ] =0.

(iii) Finally,[ϕ] =1 iffΓ`ϕ and, therefore, A_Γ(T K)has a different element of 1 iff there isϕ∈For(TK) such thatΓ0ϕ.

Naturally, if[ϕ] = 1, then Γ`ϕ and if [ϕ] =0, then Γ ` ¬ϕ. So, it results from preceding propositions that for every formulaϕ: [ϕ] =1 iffΓ`ϕ,[ϕ] =0 iff Γ` ¬ϕ, and[ϕ]6=0 iffϕis irrefutable inΓ, and, sinceA_Γ(T K)is non-degenerate, thenΓhas some non theorem.

Theorem 5.15(Soundness). The TK-algebras are correct models for the Logic TK.

Proof. LetA = (A, 0, 1,∨,∼,•)be a TK-algebra. It remains to prove that the axioms Ax_{T K1} and Ax_{T K2}are valid and the ruleR^‡preserves validity:

Ax_{T K1}: v(ϕ→‡ϕ)= v(ϕ)šv(‡ϕ)=∼v(ϕ)∨v(‡ϕ)=∼v(ϕ)∨(v(ϕ)∨v(‡ϕ))

=(∼v(ϕ)∨v(ϕ))∨v(‡ϕ)=1∨v(‡ϕ) =1.

Ax_{T K12}: v(‡‡ϕ → ‡ϕ) = • •v(ϕ) š •v(ϕ) = ∼ • •v(ϕ)∨ •v(ϕ) = ∼ •v(ϕ)∨

•v(ϕ) =1.

R^‡: Using Proposition 2.2 (ii): v(ϕ →ψ) =1⇔v(ϕ)≤ v(ψ)⇒‡v(ϕ)≤‡v(ψ)

⇒v(‡ϕ)≤v(‡ψ)⇔v(‡ϕ→‡ψ) =1.

Theorem 5.16(Adequacy). Letϕ∈For(TK). The following assertions are equivalent:

(i) `ϕ;

(ii) ϕ;

(iii) ϕ is valid in every TK-algebra of closed subsets of a Tarski space(S,⁻);

(iv) v₀(ϕ) =1, where v₀ is the valuation defined at the canonical model.

Proof. (i)⇒(ii) it follows from Soundness Theorem. (ii)⇒(iii) it suffices to observe that the algebra of closed subsets of any Tarski space is a TK-algebra. (iii) ⇒(iv) since every TK-algebra is isomorphic to a subalgebra of closed subsets of a Tarski space(S,⁻)andA(T K)is a TK-algebra, the result follows. (iv)⇒(i) ifϕ∈For(TK) and it is not derivable in TK, by Corollary 5.12,[ϕ]6=1 inA(T K)and thusv₀(ϕ)6=

1.

Corollary 5.17(Completeness). Ifϕ∈For(TK), then: ϕ⇔ `ϕ.

In the next proposition it is proved that the formula(ϕ→ψ)→(‡ϕ →‡ψ)is not TK valid by a counter example.

Proposition 5.18. 2(ϕ→ψ)→(‡ϕ→‡ψ).

Proof. There is a TK-algebra in which does not hold the above formula. Let E = {x,y,z} and take the Boolean algebra (P(E), ^C,∩,∪,∅,E). Now, define the fol- lowing consequence operator over (P(E), ^C,∩,∪,∅,E): •{x} = {x,y}, •{x,z} = {x,y,z}, and•X=X, for all the other sets inP(E). Then(P(E), ^C,•,∩,∪,∅,E)is a TK-algebra, but the formula(ϕ→ψ)→(‡ϕ→‡ψ)is not valid in it. Below we show that v(ϕ→ψ)*v(•ϕ→ •ψ), whenϕis interpreted by{x}andψby{z}:

{x,y,z}

{x,y} {x,z} {y,z}

{x} {y} {z}

;

H HH

HH

HH

HH H

HH HH H

HH HH H

{x}š{z}={x}^C∪ {z}={y,z} ∪ {z}={y,z}and

•{x}š•{z}= (•{x})^C∪ •{z}={z} ∪ {z}={z}.

As a consequence of the previous proposition, follows that the Deduction Theo- rem is not valid for the TK Logic when it was applied the ruleR^‡in a deduction.

The next results involve the models of set of formulas with the concept of ideals in a TK-algebra.

6. Strong adequacy

Definition 6.1. LetA be a TK-algebra andΓ⊆For(TK). Analgebraic modelforΓis a valuation v: For(L)−→ A, such that for everyγ∈Γ, v(γ) =1.

We denote that A is an algebraic model for Γ ⊆ For(TK) by A  Γ, that is, A Γif, and only if, for everyγ∈Γ,v_A(γ) =1.

Definition 6.2. LetΓ∪ {γ} ⊆For(TK). The setΓimpliesγ(orγis asemantic conse- quenceofΓ) when, for every modelA, ifA Γ, thenA {γ}.

We denote thatΓimpliesγbyÏγ.

Proposition 6.3. ForΓ⊆For(TK), we have:Γ`γ⇒Ïγ.

Proof. Letv:For(L)−→ A be a model forΓ. SinceA is a TK-algebra, thenv is a model for every axiom of TK and for everyγ∈Γ. As in the Soundness Theorem, the rules of TK preserve validity and if Γ`γ, then v_A(γ) =1.

Definition 6.4. LetΓ⊆ For(T K). A model v: For(TK)−→ A isstrongly adequate forΓwhen, for everyγ∈For(TK):

Γ`γ⇔Ïγ.

Proposition 6.5. LetΓ⊆For(TK) be consistent. Then:

(i) the algebraA_Γ(T K)is non degenerate;

(ii) the canonical valuation v₀ is an adequate model for Γ in A_Γ(T K) such that Γ`σ⇔if v₀(γ) =1, for everyγ∈Γ, then v0(σ) =1.

Proof. (i) AsΓis consistent, there isσsuch thatΓ0σ. So[σ] =0∈ A_Γ(T K)and for any axiom ϕ of TK it follows that[ϕ] = 1 ∈ A_Γ(T K). Then A_Γ(T K) is non degenerate.

(ii)A_Γ(T K)is a TK algebra and by construction v₀= [ ]andΓ`γ⇔[γ] =1, for anyγ∈Γ⇒[σ] =1.

Theorem 6.6. LetΓ⊆For(TK). The following conditions are equivalent:

(i) Γis consistent;

(ii) there is an adequate model forΓ;

(iii) there is an adequate model for Γ in a TK-algebra A of all closed subsets of a Tarski space(S,⁻);

(iv) there is a model toΓ.

Proof. (i)⇒(ii)It follows of preceding proposition.

(ii)⇒(iii)SinceA_Γ(T K)is a TK-algebra and every TK-algebra is isomorphic to a subalgebra of closed sets of a Tarski space(S,⁻), then the result follows.

(iii)⇒(i v)It is an immediate consequence.

(vi)⇒(i)LetA be a model forΓand suppose thatΓis not consistent. ThenΓ`γ and Γ` ¬γ. So v_A(γ) = 1 and v_A(¬γ) =1= ∼v_A(γ). But if v_A(γ) = 1, then

∼v_A(γ) =0 and therefore we have a contradiction.

Corollary 6.7. LetΓ∪ {γ} ⊆For(T K)consistent. The following conditions are equi- valent:

(i) Γ`γ; (ii) Ïγ;

(iii) every model ofΓin a TK-algebra of all closed subsets of a Tarski space(S,⁻)is a model toγ;

(iv) v₀(γ) =1, for the canonical valuation v₀.

7. Some meta-theorems

In this section, some other syntactic and semantic consequent results from previous sections are established.

Theorem 7.1. Propositional calculus TK is consistent.

Proof. Suppose that TK is not consistent. Then there isϕ ∈For(TK) such that`ϕ and ` ¬ϕ. By Soundness Theorem, ϕ and ¬ϕ are valid. Let v be a valuation in a TK-algebra with two elements 2 = {0, 1}. Since ϕ is valid, then v(ϕ) = 1 and thereforev(¬ϕ) = ∼v(ϕ) =0. This contradicts the fact that¬ϕ is valid.

Proposition 7.2. LetΓ⊆For(TK). The set I_Γ={[ϕ]:∃k∈N,γ1, . . . ,γk∈Γ `ϕ→

‡(γ1∨. . . ∨γk)}is a TK-ideal in the TK-algebraA(T K).

Proof. (1)[ϕ],[ψ]∈I_Γ⇒ `ϕ→‡(γ1∨. . . ∨γk)and`ψ→‡(σ1∨. . . ∨σm)⇒

`(ϕ∨ψ)→‡(γ1∨. . . ∨γk)∨‡(σ1∨. . . ∨σm)⇒(by Proposition 4.4)`(ϕ∨ψ)→

‡(γ1∨. . . ∨γk∨σ1∨. . .∨σm)⇒[ϕ∨ψ] = [ϕ]∨[ψ]∈I_Γ.

(2)[ϕ]≤[ψ]and[ψ]∈I_Γ ⇒ `ϕ →ψ and`ψ→‡(σ1∨. . . ∨σm)⇒ `ϕ→

‡(σ1∨. . . ∨σm)⇒[ϕ]∈I_Γ.

(3) [ϕ] ∈ I_Γ ⇒ ` ϕ → ‡(γ1∨. . . ∨γk) ⇒ [ϕ] ≤ [‡(γ1∨. . . ∨γk)] ⇒ [ϕ] ≤

‡[(γ1∨. . . ∨γ_k)] ⇒‡[ϕ]≤‡[(γ1∨. . . ∨γ_k)] ⇒[‡ϕ]≤[‡(γ1∨. . . ∨γ_k)]⇒

`‡ϕ→‡(γ1∨. . . ∨γk)⇒[‡ϕ] =‡[ϕ]∈I_Γ.

In the logical context, this proposition is similar to the Proposition 3.4 about ideals. In both we could takeγ∈Γbecause ifγ1, . . . ,γk∈Γ, thenγ1∨. . . ∨γk∈Γ and also‡(γ1∨. . . ∨γk)∈Γ. So, the above TK-idealI_Γis the TK-ideal generated by {[γ]:γ∈Γ}.

The setΓ⊆For(TK) is consistent if, and only if, I_Γ is a proper ideal ofA(T K), becauseΓis not consistent iff 1∈Γiff[1]∈I_Γ iffI is not a proper ideal.

Proposition 7.3. If J is a TK-ideal inA(T K) and Γ ={ψ∈ For(T K) :[ψ]∈J}, then J ⊆I_Γ.

Proof. LetΓ ={ψ∈For(T K):[ψ]∈J}. Hence[ϕ]∈J⇒ϕ∈Γ⇒[ϕ]∈I_Γ. Proposition 7.4. LetΓ⊆For(TK). The TK-ideal I_Γ is maximal if, and only if, the setΓ is maximal

Proof. (⇒)Suppose that Γis not maximal. In this case, there isψ∈For(TK) such thatΓ1= Γ∪ {ψ}andΓ2= Γ∪ {¬ψ}are consistent. SoΓ⊆Γ1∩Γ2, withΓ16= Γ2. Therefore, I_Γ ⊆ I_Γ1∩I_Γ2 and since I_Γ is maximal, then I_Γ = I_Γ1 = I_Γ2 what is a contradiction, because[ψ]∈I_Γ1,[¬ψ]∈I_Γ2and 1∈/I_Γ.

(⇐)IfΓis maximal, for eachψ∈For(TK), eitherψ∈Γor¬ψ∈Γ. Then for each ψ∈For(TK),[ψ]∈I_Γ or[¬ψ]∈I_Γ. SinceΓis consistent, thenI_Γ is proper and it is not the case that 1=[ψ]∨[¬ψ]∈I_Γ. Hence, the TK-idealI_Γis maximal.

Proposition 7.5. If Γ⊆For(TK) is consistent, then there is a maximal set∆⊆For(TK) such thatΓ⊆∆.

Proof. The result follows from Zorn’s Lemma.

Proposition 7.6. If Γ⊆ For(TK) is consistent, then there is∆⊆For(TK) irreducible such thatΓ⊆∆.

Proof. IfΓis consistent, by previous proposition, there is a maximal set∆such that Γ⊆∆. Now, since each maximal set is an irreducible set, then∆is irreducible.

Definition 7.7. A model for Γ in a non degenerate TK-algebra A is a•-semantic modelofΓwhen for everya∈A: either•a=1 or•a=0.

Proposition 7.8. If Γ⊆For(TK) is consistent, thenΓhas a•-semantic model.

Proof. IfΓis consistent, it is included in a maximal setΓ^∗and it has a modelA. By Proposition 7.4, I_Γ∗ is maximal and from Proposition 3.24,A |J^∗ is a non degenerate TK-algebra with the property that for every ¯a ∈ A |J^∗, either •¯a ≡ 0 or •¯a ≡ 1.

If his the surjective homomorphismh:A −→ A |J^∗, so the compositionhov is a

•-semantic model forΓ.

GivenΓ⊆ For(TK), consider the functionh:A(T K)−→ A_Γ(T K), defined by h([ψ]) = [ψ]_Γ. We are going to denote a member ofA(T K)by[ψ], and a member ofA_Γ(T K)by[ψ]_Γ. Naturallyhis a surjective homomorphism. Now, the kernel of his:

Ker(h) = {[ψ] ∈ A(T K) : h([ψ]) = [0]_Γ} = {[ψ]∈ A(T K) : [ψ]_Γ = [0]_Γ} = {[ψ]∈ A(T K):Γ` ¬ψ}, because[ψ]<sub>Γ</sub>= [0]<sub>Γ</sub> ⇔Γ`ψ→0 and Γ`0→ψ⇔ Γ` ¬ψandΓ`1⇔Γ` ¬ψ.

By Theorem 3.22,A(T K)|Ker(h)is isomorphic toA_Γ(T K)

Theorem 7.9. GivenΓ⊆For(TK), the following statements are equivalent:

(i) for everyϕ∈For(TK) exactly one holds: Γ`‡ϕorΓ` ¬‡ϕ;

(ii) Γis maximal;

(iii) A_Γ(T K)is isomorphic to a non degenerate TK-algebra of sets (Tarski space) A which is a•-semantic model ofΓ;

(iv) Γis consistent and each•-semantic model forΓis adequate.

Proof. (i)⇒(ii) Suppose that Γis not maximal. So there is ∆ maximal such that Γ ⊂ ∆. From Proposition 7.5, there is [ψ] ∈ I_∆ but [ψ] ∈/ I_Γ. Then [‡ψ] ∈/ I_Γ and Γ0 ‡ψ. By (i) Γ ` ¬‡ψ, then ∆ ` ¬‡ψ and∆ `‡ψ, what contradicts the maximality of∆.

(ii)⇒(iii) By previous analysisA_Γ(T K)≈ A(T K)|Ker(h). Now, sinceΓis maximal, A_Γ(T K) is not degenerate and, for every ψ ∈ For(TK), Γ ` ψ ⇔ ψ ∈ Γ. So, for every ψ ∈For(TK), either [‡ψ] = 0 or [‡ψ] = 1, that is, A(T K)|_Ker(h) is•- semantic. From Theorem 2.10, this TK-algebra is isomorphic to a TK-algebra of sets (or Tarski space)A.

(iii) ⇒ (iv) Considering (iii), the set Γ is consistent. Now let A be an arbitrary

•-semantic model for Γ. Given ϕ ∈ For(TK), either v_A(ϕ) = 0 or v_A(ϕ) = 1. If v_A(ϕ) =1, thenΓ`‡ϕand ifv<sub>A</sub>(ϕ) =0, thenΓ0 ‡ϕ, and thusΓ` ¬‡ϕ.

(iv) ⇒(i) SinceΓis consistent, by Proposition 7.8, there is a•-semantic modelA for it. By (iv) this model is adequate. So for anyϕ∈For(TK), eitherv_A(ϕ) =0 or v_A(ϕ) =1, that is, eitherΓ`‡ϕ orΓ` ¬‡ϕ.

Corollary 7.10(Decidability). The Logic TK is decidable.

Proof. Consider the•-semantic model2={0, 1}, such that•0=0 and•1=1. Since from previous theorem any •-semantic model is adequate, therefore 2is adequate and, in this way, for any formulaψ∈For(TK),`ψ⇔v₂(ψ) =1.

8. Final considerations

Logic TK is a kind of modal logic. The operator‡has an intuitive algebraic interpre- tation and another one given by the Tarski spaces.

As it is known from the modal logics, in the presence of the necessitation rule (NR - Proposition 4.3) the K axiom: ‡(ϕ → ψ) → (‡ϕ → ‡ψ) is equivalent to

‡(ϕ∧ψ) ↔ (‡ϕ∧‡ψ). However, only ‡(ϕ∧ψ) → (‡ϕ ∧‡ψ) holds in TK.

Therefore we can observe that TK is a subnormal modal logic. It is easier to see that by analyzing the dual operator◊.

Of course, we can investigate other kind of semantics for TK, particularly, some relational semantic kind.

Maybe some variations on TK-algebras can provide natural and simple algebraic models to other modal logics.

Acknowledgements

This work has been sponsored by FAPESP.

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HÉRCULES DEARAÚJOFEITOSA

São Pauko State University – UNESP/FC

BRAZIL

haf@fc.unesp.br MAURICUNHA DONASCIMENTO

São Pauko State University - UNESP/FC

BRAZIL

mauri@fc.unesp.br MARIACLAUDIACABRINIGRÁCIO

São Pauko State University - UNESP/FC

BRAZIL

cabrini@marilia.unesp.br Resumo. Tarski apresentou sua definição de operador de consequência com a intenção de expor as concepções fundamentais da consequência lógica. Um espaço de Tarski é um par ordenado determinado por um conjunto não vazio e um operador de consequência sobre este conjunto. Esta estrutura matemática caracteriza um espaço quase topológico. Este artigo mostra uma visão algébrica dos espaços de Tarski e introduz uma lógica proposicional modal que interpreta o seu operador modal nos conjuntos fechados de algum espaço de Tarski.

Palavras-chave:Espaço de Tarski, espaço quase topológico, operador de consequência, ló- gica modal, modelo algébrico.